SWIG CYGWIN DLL链接 [英] SWIG CYGWIN DLL linking

问题描述

我想按照这里的SWIG Java范例。这个例子使用cygwin进行编译。我想传递编译器一个alredy编译dll，test.dll，而不是一个C源文件。 SWIG.org示例使用C源。我试图传递test.dll（编译C源），然后example.dll到第三个命令，但我得到一个错误test.dll：没有这样的文件或目录。 SWIG.org示例的第3个命令基于example.c创建example.dll。如何创建example.dll，以便它使用test.dll而不是example.c？

I'm trying to follow the SWIG Java example located here. This example uses cygwin for compilation. I would like to pass the compiler an alredy compiled dll, test.dll, instead of a C source file. The SWIG.org example uses C source. I attempted to pass the test.dll(compiled C source) and then example.dll to the 3rd command but I get an error "test.dll: no such file or directory". The SWIG.org example's 3rd command creates the example.dll based on example.c. How can I create example.dll so that it uses test.dll instead of example.c?

有没有办法在此范例的上下文中完成此操作？

Is there a way to accomplish this within the context of this example?

C Source，just dll：

$ swig -java example.i

$ swig -java example.i

$ gcc example_wrap.c -I / c / jdk1.6.0_30 / include -I / c / jdk1.6.0_30 / include / win32

$ gcc example_wrap.c -I/c/jdk1.6.0_30/include -I/c/jdk1.6.0_30/include/win32

$ gcc -shared example_wrap.o -mno-cygwin -Wl， - add-stdcall-alias -o test.dll example.dll

$ gcc -shared example_wrap.o -mno-cygwin -Wl,--add-stdcall-alias -o test.dll example.dll

SWIG.org示例代码：

$ swig -java example.i
$ gcc **-c example.c** example_wrap.c -I/c/jdk1.3.1/include -I/c/jdk1.3.1/include/win32
$ gcc -shared example.o  example_wrap.o -mno-cygwin -Wl,--add-stdcall-alias  -o example.dll

推荐答案

您需要将 test.dll 传递给最后一个命令，该命令链接您的程序，而不是第二个命令，它只是编译了example_wrap.c源文件。 （ -c 选项指示gcc只能编译。）

You need to pass test.dll to the last command, which links your program, rather than the second, which just compiles the example_wrap.c source file. (The -c option tells gcc to compile only.)

请注意，对象和DLL在链接线很重要。 DLL应该在使用它们的任何东西之后。

Note that the ordering of objects and DLLs on the link line is important. DLLs should come after anything that uses them.

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